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1st problem's all good
But for 2nd problem :
Ff =-7.6 N Thats cool but his weight is going down the hill with a force that would be equal to +10 N if there was no friction ->Fw= 3kg*9,81*sin(20)=10N
so global force applied on this block is f=10N-7,6N=2,4N
And work should be : W=2,4N*1,5m=2,6J
If you want to know the work done by friction then it should be w = (7.6)(1.5)cos 180=-11,4J (The - sign mean the work's done in a direction opposed to your references axes (we suppose the X axe is growing in the direction the weight's moving)
and if you want the work done by the weight it should be w = 10N*1,5*cos(0)=15J ... 15J-11,4J=2,6J
Im not rly shure tho.. im confused with this old definition of work.. but w-e the overall energy lost by the block will be 11,4J, the overall potential energie lost by the block is 15J. The 2,6J of energy left at the end is cinetic energie.. the potential energy was either lost due to friction or transformed in cinetic energie (mouvement).
Later knowing your block gained 2,6J or cinetic energie you can determin his speed at the bottom of that 1,5m hill will be 1,32m/s with
2,6J=(3*v^2)/2 -> v=(5,2/3)^1/2=sqrt(5,2/3)=1,32 m/s